How many moles of oxygen are required to react with 18.3 mol ammonia? What's the equation and how's it done?

Question

I need help with setting up the equation,,,,,,,,,

Answer 1

The normal oxidation of NH3 gives N2 and H2O so it has to be specially passed over a Pt gauze at 850ºC:

4 NH3 + 5 O2 → 4 NO + 6 H2O

ref: http://en.wikipedia.org/wiki/Ammonia

18.3 moles requires 5/4 times as much O2 or 22.875 moles

If you want the normal combustion reaction it's

4NH3 + 3O2 --> 2N2 + 6H2O

in which case you'd need 3/4 moles O2 (13.725 moles)

Answer 2

The chemical equation without any coefficients is

NH3 + O2 ==> NO + H2O

Balanced, it becomes

4 NH3 + 5 O2 ==> 4 NO + 6 H2O

You can see that you need 5 mol of oxygen gas to react with 4 mol of ammonia, so you need 1.25 times as many mol of O2 as of NH3. If you heave 18.3 mol of NH3, you need 1.25*18.3 = 22.875 mol of O2.